What is the integral from 1 to 2 of (1+e^x) ?

The integral from 1 to 2 of (1+e^x) is 1+e^2-e

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integral from 1 to 2 of (1+e^x)

\int_{1}^{2} (1 + e^{x}) d x

1 + e^{2} - e

Decimal

5.67077 \dots

Solution steps

\int_{1}^{2} 1 + e^{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} 1 d x + \int_{1}^{2} e^{x} d x

\int_{1}^{2} 1 d x = 1

\int_{1}^{2} e^{x} d x = e^{2} - e

= 1 + e^{2} - e

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What is the integral from 1 to 2 of (1+e^x) ?
The integral from 1 to 2 of (1+e^x) is 1+e^2-e